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Prove that if n is a perfect square, then n +2 is not a perfect square

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i suppose first step would be to assume n is perfect square
so n = m^2
then i suppose i find n^2

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i mean n + 2
n + 2 would be m^2 + 2
so now... i assume m^2 + 2 is a perfect square...
we have no 2 perfect square numbers with difference 2
so i let m^2 + 2 = k^2
@mukushla that's not really a proof.....
you may continue @igba
hmm i suppose the next step would be m^2 - k^2 = 2 so (m+k)(m-k) = 2 i don't think this proves anything...
actually that is :) \[m^2-k^2=2\]\[(m-k)(m+k)=2\]and this is impossible
why so?
well,,you've almost proved it.. now m and k both are integers/.
and its k^2 -m^2 = 2 and not otherwise..
oh yes
2 = (k + m)(k-m)
now...integers huh...
\[m+k>m-k\]so\[m+k=2\]so\[m=k=1\]but it gives\[m-k=0\]
you know k>m.. so from (k+m)(k-m)= 1*2 =>k+m =2 and k-m = 1 this doesnt have an integral solution..
why do i know k > m again?
m^2 + 2 = k^2 both m and k are positive integers hence k>m
in layman language,, k^2 - m^2 >0
when was it assumed that m and k are positive?
what else do you mean by perfect square ?
i don't see how that relates.... the definition of a perfect square is x = k^2
4 is a perfect sq since its 2^2 i.e. square of a positive integer.. and likewise..
@mukushla "we have no 2 perfect square numbers with difference 2" Because 0,1,4,9,6,5,6,9,4,1 etc....
(-2)^2 is also 4...
you take the absolute value
i mean you have to..
see,, (k+m)(k-m) =2 either both are positive are both are negative,,since RHS is postive.. so even if you take -ve values,, negative * negative = positive ,,i.e. treated as absolute value..
...i don't think this is a valid proof....
another explanation might me x is a perfect square because its y^2 x = y^2 = (-y)^2 sqrt(x) = | y | = | -y |
who can writing here now two perfect squares with difference of 2 ? so because i think do not exist or .... ???
since when did sqrt x become |-y|
0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats. No +2 in there.
so and i think that this difference of two will be the ,,key" of this proof of ...
@lgbasallote what is your opinion from this ,please ?
my opinion is that this is some tricky algebra
how do you think it ???
so and why ?
because i can't think of a proof
how can you prove/disprove that the difference of two squares is 2...
so i think this is very easy logicaly but to prove it will be difficile i think so but there are again @saifoo.khan and @satellite73 probably they will can doing it hope so much
their fields of specialty aren't exactly in proving
What's wrong with my proof?
anyway... m^2 + 2 = k^2 k^2 - m^2 = 2 for it to be equal to 2...k^2 and m^2 should either be both even...or both odd so if i assume both are even.. then k and m are also both even so i can rewrite this as (2x)^2 - (2y^2) = 2 4x^2 - 4y^2 = 2 then... 2x^2 - 2y^2 = 1 x^2 - y^2 = 1/2 then since it's x^2 and y^2 are both integers...the difference should also be integer \('\therefore\) contradiction
and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..
@estudier i mean
So, that doesn't mean it's invalid..
it does can't do infinite substitutions to prove anything....
There is no infinity involved, all integer squares end in the numbers I gave.
infinite loops involves infinity agree?
No infinite loop either, just logic.
An integer ends in one of digits 0 to 9. The squares must end in etc....
but we both know that setting values is illegal in proving
I am not setting any values.
and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever
you're checking the squares of each number
I am not.
and though there is a pattern...there is still susbtitution involved
and then you'll have to substitute forever to verify that the pattern really does not break
then infinity is involved
It's not a pattern, it's a fact 0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1
so you do admit that you checked the 0 to 9 squares
Anything else is impossible.
...that's substitution point is...substitution isn't really allowed in proving
You will find that this fact is used a lot in many valid proofs in number theory.
OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.
since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?
I am not using brute force, brute force is something like testing or experimental math. Or exhaustion (which is also a valid proof technique). It's the use of a fact about squares (ALL squares)
can you prove it? that it's a fact about squares?
Yes, all numbers end in 0 to 9 and so their squares end in .....QED.
you didn't really demonstrate just repeated what you said
You asked for proof, I just gave u a proof.
If you deny my proof, provide a counterexample.
since you try so hard to defend your proof, i'll accept it
That's very gracious, thank you:-)
do check on my algebra proof though please. i would like to know if that's the way contradiction works
OK, I will look at it now....
Did u do the case when both are odd? (Perhaps a more straightforward way is one you start with n= some k squared then next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)
why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?
No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)
oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement
again i ask... how is k > m
(k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
factoring involves lesser steps...its obvious that k>m \[k^2−m^2=2\]\[(k−m)(k+m)=2\]easily u can see \[k-m=1\]\[k+m=2\]add them\[2k=3\] Contradiction
again i ask for the nth time @mukushla how is k > m
ok man i just want to answer ur question
"2 can only be an outcome of even and even or odd and odd" is a valid statement Why do you think it might not be valid?
well because 8 - 4 is not 2 but it's even minus even
@mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?
OK, so finally, the algebra Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
i thought they are positive :) sorry
why (k+1)^2 @estudier ?
I already, said, that's the next square....
next square?
is this still related to my proof?
? n is k^2, what is the next square?
no problem ... this is better\[|k|^2-|m|^2=2\]\[(|k|-|m|)(|k|+|m|)=2\]so\[|k|-|m|=1\]\[|k|+|m|=2\]because this time im sure that \(|k|>|m|\) add them\[2|k|=3\]Contradiction and we are done
No, this proof Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
are you allowed to do that? @mukushla
can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end
why not...for any real number \[a^2=|a|^2\]
Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 k^2 + 2k +1 >= n +3 n +3 > n+2 QED
That's 2 proofs I have given now...
a few questions in mind @estudier 1) why find the next square? 2) how is k^2 + 2k + 1 > = n + 3 3) what does n + 3 > n+2 prove?
Think about it....:-)
in the meanwhile... @mukushla why |k| - |m| = 1?
how does that make |k| - |m| = 1
u have 2 positive integers with ab=2 and a
who said those are positive?
|k|>|m| right?
something feels wrong...
anyway...where did you get 2|k| > 3
ok...think on what i said...because i believe thats the shortest way to prove ur statement
|2k| = 3 *
Going back to your question about 2, u can just adjust it to difference of 2q...
im really skeptical about this... especially that |k|^2 - |m|^2 = 2
difference of 2q?
You were worried about the validity of your statement, change it to 2q instead of just 2.
Or 2p, or whatever letter u like best.
1) why find the next square? Because we want to see that it isn't n+2 2) how is k^2 + 2k + 1 > = n + 3 Substitute n=k^2 3) what does n + 3 > n+2 prove? See 1)
oh. when i said difference of two numbers equal to 2
but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?
and @mukushla how did you get |k| > |m|
No, because the next square is (k+1)^2 (which might or might not be equal to n+2, that is what we have to discover.
and since the next square is greater than n + 2...that makes n + 2 invalid for any succeeding squares?
Yes, the algebra is completely general...
i'm still confused about k^2 + 2k + 1 though...if n = k^ just makes n + 2k + 1...
...which is greater than n+3, right?
it is?
Number theorists don't pay attention to 0
unless k is 0....
why not?
Tradition, convention, call it as u like..... So in fact there are 2 definitions for natural number , one with 0 and one without.
Can I try? give so many weird proofs......the sad part is that i cannot disprove it because i know nothing about it....
More the merrier.
"because i know nothing about it" But you do now:-)
it's not like it's limited to the 99s....
i was referring to your claims earlier like brute force is valid proof in number theory and that number theorists don't consider 0...i cannot disprove your proofs because i do not know if these claims are true or not
Number theory is less onerous than real number stuff because it is integer only. But it is actually this difference that causes all the difficulty, students are not accustomed to thinking in a more freewheeling way.
Let n be a perfect square where n=m^2 then next consecutive perfect square number would be (m+1)^2 Now, (m+1)^2-m^2=2m+1 Since, 2m+1>2 as m is a set of natural numbers. Thus, if n is a perfect square..... n+2 cannot be a perfect square as the difference between two perfect squares is always more than 2.
Yes, that's similar to the one I gave...
"number theorists don't consider 0" Using 0,1,4, etc...
ok,so who can answering this question ? ,,the line of perfect squares begin with number ... " what is this number ?
so and for this your question @lgbasallote i think that for n=m^2 and n+2 for being a perfect square so than n+2=m^2 +2 suppose m^2 +2 =x^2 like a perfect square so than m^2 = x^2 -2 so m1,2 = +/- sqrt(x^2 -2)

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